是否可以将动态框架转换为静态库？
问题是我没有框架的源代码，只有<?php
// This scripts sets up PHP environment variables to display all errors on
the screen
// and then connects to an oracle database
// Send PHP errors to the screen (should always be off for live systems)
ini_set('display_errors', 1);
// Set PHP error reporting level to report all errors.
error_reporting(E_ALL);
echo "<style type='text/css'> p {font-size: x-large} </style>";
// CONNECT TO ORACLE STUDENT ACCOUNT AT HALLAM
// Display 'connecting' message
echo "<p>Connecting . . . </p>";
// Set up variables for connection
$con_hostname = 'homepages.shu.ac.uk:1521/shu11g.shu.ac.uk';
// CHANGE THE LINES BELOW TO ADD YOUR LOGIN DETAILS !!!
$con_username = 'B009';
$con_password = 'J123df';
// Connect to the database. $conn is a variable of type 'resource', and is
the connection handle.
// It returns a connection identifier, or FALSE (if it fails to connect)
$conn = oci_connect($con_username, $con_password, $con_hostname);
if (!$conn) {
$e = oci_error();
trigger_error(htmlentities($e['message'], ENT_QUOTES), E_USER_ERROR);
}
else { echo "<p> . . . Connected</p>";}
?>
文件，这是少数架构的动态库？
答案 0 :(得分：1)
创建“ Cocoa Touch Framework ”并将您的.framework添加到“链接的框架和库”，然后您可以：
或
不要忘记添加modulemap / Umbrella以展示您想要的内容。